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I am considering the power series $$\sum\limits_{k=1}^{\infty} \frac{x^k}{k}$$ and its convergence. I checked easily by Ratio test that it converges absolutely for $|x| < 1$ and I check the endpoints $-1,1$. In the first case I got $\sum\limits_{k=1}^{\infty} \frac{(-1)^k}{k}$ which converges, but not absolutely. For the other case, we have the divergent harmonic series. So actually my series converges on $[-1,1)$.

For uniform convergence is it true that it converges uniformly in $[-s,s]$ for $s \in [0,1)$ but not on $(-1,1)$ because at $-1,1$ we did not have absolute convergence?

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At $1$ it diverges and therefore it could not possibly converg uniformly on any set to which $1$ belongs.

And there is no connection between uniform convergence and absolute convergense. It converges uniformly on $[-s,s]$ by the Weierstrass $M$-test. In fact, it converges uniformly on any set of the form $[-1,s]$, with $0<s<1$, by Dirichlet's test.

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  • $\begingroup$ Ok for $1$ it makes sense. Using the M-test what series should I compare this to? $\endgroup$ – mandella Apr 23 '18 at 14:20
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    $\begingroup$ @mandella You should compare it to $\frac{s^n}n$. $\endgroup$ – José Carlos Santos Apr 23 '18 at 14:24
  • $\begingroup$ And this one will converge because of the ratio test since $|s| < 1$ correct? $\endgroup$ – mandella Apr 23 '18 at 14:25
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    $\begingroup$ @mandella Indeed. $\endgroup$ – José Carlos Santos Apr 23 '18 at 14:25
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    $\begingroup$ @mandella Because bounded functions never converge uniformly to an unbounded function. $\endgroup$ – José Carlos Santos Apr 23 '18 at 14:41

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